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Chapter 12Machine Learning
ID: 116 117Name: Qun Yu (page 1-33) Kai Zhu (page 34-59)Class: CS267 Fall 2008Instructor: Dr. T.Y.Lin
Introduction
Machine Learning is a key part of A.I. research.
Rough Set Theory can be used for some problems in Machine Learning.
Will discuss 2 cases:1) Learning from Examples2) An Imperfect Teacher
Case1:Learning from Examples
We assume: 2 agents: a knower, a learner Knower’s Knowledge set U U is unchanged, will NOT increase during the
learning process. It is called CWA(Closed World Assumption)
Knower knows everything about U Learner has ability to learn U, in other words,
learner knows some attributes of objects in U
Example For Case 1 KR-System
U a b c d e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
4 0 0 1 2 1
5 2 1 0 2 1
6 0 0 1 2 2
7 2 0 0 1 0
8 0 1 2 2 1
9 2 1 0 2 2
10 2 0 0 1 0
Attributes of learner’s knowledge : B={a,b,c,d}
Attributes of knower’s knowledge : e
Example For Case 1 KR-System
U a b c d e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
4 0 0 1 2 1
5 22 11 00 22 1
6 0 0 1 2 2
7 2 0 0 1 0
8 0 1 2 2 1
9 22 11 00 22 2
10 2 0 0 1 0
3 concepts of knower’s knowledge:
X0={3,7,10}
X1={1,2,4,5,8}
X2={6,9}
5 concepts of learner’s knowledge:
Y0={1,2}
Y1={3,7,10}
Y2={4,6}
YY3={5,9}={5,9}
Y4={8}
Which Objects are learnable?X0={3,7,10}
X1={1,2,4,5,8}
X2={6,9}
Y0={1,2}
Y1={3,7,10}
Y2={4,6}
Y3={5,9}
Y4={8}
BX0 = Y1 = {3,7,10} = X0 = BX0
X0 is exactly Y-definable and can be learned fully.
BX1 = Y0 Y4 = {1,2,8}
BX1 = Y0 Y2 Y3 Y4 = {1,2,4,5,6,8,9}
X1 is roughly Y-definable, so learner can learn object {1,2,8}, not sure about {4,5,6,9}.
BX2 = Ø
BX1 = Y2 Y3 = {4,5,6,9}
X2 is internally Y-indefinable, so it is NOT learnable.
Learnable:Learnable: The knower’s knowledge can be expressed in terms of learner’s knowledge.
Quality of Learning?BX0 = Y1 = {3,7,10} = X0 = BX0
BX1 = Y0 Y4 = {1,2,8}
BX1 = Y0 Y2 Y3 Y4 = {1,2,4,5,6,8,9}
BX2 = Ø
BX1 = Y2 Y3 = {4,5,6,9}
X0 X1 X2
Positive objects 3,7,10 1,2,8 Ø
Border-line objects Ø 3,7,10 4,5,6,9
Negative objects 1,2,4,5,6,8,9 4,5,6,9 1,2,3,7,8,10
POSB{e}=POSB(X0) POSB(X1)
POSB(X2) = {1,2,3,7,8,10} (6 objects)
U={1,2,3,4,5,6,7,8,9,10} (10 objects)
{e}= 6/10 = 0.6
Simplifier table…step 1U a b c d e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
4 0 0 1 2 1
5 22 11 00 22 1
6 0 0 1 2 2
7 2 0 0 1 0
8 0 1 2 2 1
9 22 11 00 22 2
10 2 0 0 1 0
Group the duplicate rows :
U a b c d e
3,7,10
2 0 0 1 0
1,2 1 2 0 1 1
4 0 0 1 2 1
5 22 11 00 22 1
8 0 1 2 2 1
6 0 0 1 2 2
9 22 11 00 22 2
Simplifier table…step 2Remove inconsistent rows :
U a b c d e
3,7,10
2 0 0 1 0
1,2 1 2 0 1 1
8 0 1 2 2 1
U a b c d e
3,7,10
2 0 0 1 0
1,2 1 2 0 1 1
4 0 0 1 2 1
5 22 11 00 22 1
8 0 1 2 2 1
6 0 0 1 2 2
9 22 11 00 22 2
Simplifier table…step 3
{a}, {b}
Find attributes reduct :U a b c d e
3,7,10
2 0 0 1 0
1,2 1 2 0 1 1
8 0 1 2 2 1
U b c d e
3,7,10
0 0 1 0
1,2 2 0 1 1
8 1 2 2 1U a c d e
3,7,10
2 0 1 0
1,2 1 0 1 1
8 0 2 2 1
U a b d e
3,7,10
2 0 1 0
1,2 1 2 1 1
8 0 1 2 1U a b c e
3,7,10
2 0 0 0
1,2 1 2 0 1
8 0 1 2 1
Delete a:
Delete b:
Delete c:
Delete d:
Simplifier table…step 4Find value reduct & decision rules:
U a e
3,7,10
2 0
1,2 1 1
8 0 1
U b e
3,7,10
0 0
1,2 2 1
8 1 1a2e0
a1e1
a0e1
a2e0
a1Va0e1
b0e0
b2e1
b1e1
b0e0
b2Vb1e1
All objects are necessary in learning process?
From the KR-System table, we can see some objects are same by attributes {a,b,c,d,e}:
1 = 2
3 = 7 =10
D1={1,2}
D2={3,7,10}
Therefore, only one object, either 1 or 2 in D1 is necessary in learning process. So is D2.
4,5,6,8,9 are necessary.
U a b c d e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
4 0 0 1 2 1
5 2 1 0 2 1
6 0 0 1 2 2
7 2 0 0 1 0
8 0 1 2 2 1
9 2 1 0 2 2
10 2 0 0 1 0
Prove from Decision Rule…(1)
U a b c d e
1 1 2 0 1 1
3 2 0 0 1 0
4 0 0 1 2 1
5 22 11 00 22 1
6 0 0 1 2 2
8 0 1 2 2 1
9 22 11 00 22 2
Remove objects 2,7 and 10 (keep 1 and 3) as Table1,Table1 will provide the same decision rules.
Table1
a2 e0
a1 e1
a0 e1
b0 e0
b2 e1
b1 e1
OR
Prove from Decision Rule…(2)
U a b c d e
2 1 2 0 1 1
4 0 0 1 2 1
5 22 11 00 22 1
6 0 0 1 2 2
8 0 1 2 2 1
9 22 11 00 22 2
10 2 0 0 1 0
Remove objects 1,3 and 7 (keep 2 and 10)as Table 2,Table2 will provide the same decision rules.
Table 2
a2 e0
a1 e1
a0 e1
b0 e0
b2 e1
b1 e1
OR
Prove from Decision Rule…(3)
U a b c d e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
5 22 11 00 22 1
6 0 0 1 2 2
7 2 0 0 1 0
9 22 11 00 22 2
10 2 0 0 1 0
Remove objects 4 and 8 as Table3,
The whole new decision algorithm will be:
Table 3
a2 e0
a1 e1
a0 e2
Now concept X2 Now is internally definable. Object 6 is learnable now. So decision algorithm changed.
b2c0 e1
b0c0 e0
b0c1 e2
b2d1 e1
b0d1 e0
b0d2 e2
OR OR
Prove from Decision Rule…(4)
U a b c d e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
4 0 0 1 2 1
5 22 11 00 22 1
6 0 0 1 2 2
7 2 0 0 1 0
8 0 1 2 2 1
10 2 0 0 1 0
Table 4
Remove objects 9 as Table 4,
The whole new decision algorithm will be:
b0 e0
b2 e1
a0b1 e1
a2b1 e1Now object 5 is positive object of X1. So decision algorithm changed.
a2d1 e0
a1d1 e1
a2d2 e1
a0d2 e1
OR
Case2: An Imperfect TeacherWe assume:
2 agents: a knower, a learner
Knower’s Knowledge set U
U is unchanged, will NOT increase during the learning process. It is called CWA(Closed World Assumption)
Knower knows everything about U
Learner has ability to learn U, in other words, learner knows some attributes of objects in U
Example 1 For Case 2 KR-System
U a b c
1 0 2 +
2 0 1 +
3 1 0 +
4 0 1 0
5 1 0 0
6 1 1 -
7 2 1 -
8 0 1 -
9 1 0 -
Attributes of learner’s knowledge : B={a,b}
Attributes of knower’s knowledge : c
Attribute value 0: knower can’t classify this object.
Example 1 For Case 2KR-System
3 concepts of knower’s knowledge:
X0={4,5} ignorance region
X+={1,2,3}
X-={6,7,8,9}
X*=X+ X- competence region5 concepts of learner’s knowledge:
Y0={1}
Y1={2,4,8}
Y2={3,5,9}
YY3={6}={6}
Y4={7}
U a b c
1 0 2 +
2 0 1 +
3 1 0 +
4 0 1 0
5 1 0 0
6 11 11 -
7 2 1 -
8 0 1 -
9 1 0 -
Can the learner be able to discover the ignorance region?
BX+ = Y0= {1}
BX+ = Y0 Y1 Y2 = {1,2,3,4,5,8,9}
X+ is roughly Y-definable, so learner can learn object {1}, not sure about {2,3,4,5,8,9}.
BX- = Y3 Y4 = {6,7}
BX- = Y1 Y2 Y3 Y4 = {2,3,4,5,6,7,8,9}
X- is roughly Y-definable, so learner can learn object {6,7}, not sure about {2,3,4,5,8,9}.
BX0 = Ø
BX0 = Y1 Y2 = {2,3,4,5,8,9}
X0 is internally Y-indefinable, so it is NOT learnable.
X0={4,5} ignorance region
X+={1,2,3}
X-={6,7,8,9}
X*=X+ X- competence region
Y0={1}
Y1={2,4,8}
Y2={3,5,9}
Y3={6}
Y4={7} Learner can’t discover the ignorance region X0
The ignorance region influences the learner’s ability to learn?
In this example, the answer is NO, because the ignorance region(X0) is not learnable.
Hence, we can prove it:
BNB(X*)= BNB(X+) BNB(X-)
BX+ = Y0= {1}
BX+ = Y0 Y1 Y2 = {1,2,3,4,5,8,9}
BX- = Y3 Y4 = {6,7}
BX- = Y1 Y2 Y3 Y4 = {2,3,4,5,6,7,8,9}
X*=X+ X- competence region
X0={4,5} ignorance regionX* X*
put 4 into X+
X*put 4 into X-
Positive objects
1,6,7 1,6,7 1,6,7
Border-line objects
2,3,4,5,8,9 2,3,4,5,8,9 2,3,4,5,8,9
Negative objects
Ø Ø Ø
Border-line region of the competence region X* remain unchanged, therefore, it doesn't matter knower knows it or NOT.
Put 4 into X+
X+={1,2,3,4}
X-={6,7,8,9}
X0={5}
BX+ = Y0= {1}
BX+ = Y0 Y1 Y2 = {1,2,3,4,5,8,92,3,4,5,8,9}
BX- = Y3 Y4 = {6,7}
BX- = Y1 Y2 Y3 Y4 = {2,3,4,52,3,4,5,6,7,8,98,9}
BNB(X*)= BNB(X+) BNB(X-)
={2,3,4,5,8,9} {2,3,4,5,8,9}
={2,3,4,5,8,9}
X0={4,5} ignorance
region
X+={1,2,3}
X-={6,7,8,9}
X*=X+ X- competence
region
Y0={1}
Y1={2,4,8}
Y2={3,5,9}
Y3={6}
Y4={7}
Calculation Sheet:Border-line objects
Simplifier table…step 1Remove inconsistent rows :
U a b c
1 0 2 +
6 1 1 -
7 2 1 -
U a b c
1 0 2 +
2 0 1 +
3 1 0 +
4 0 1 0
5 1 0 0
6 11 11 -
7 2 1 -
8 0 1 -
9 1 0 -
Learner can’t discover the ignorance region {4,5} and the ignorance region won’t affect the learning process.
Simplifier table…step 2
{a}, {b}
Find attributes reduct :
Delete a:
Delete b:
U a b c
1 0 2 +
6 1 1 -
7 2 1 -U a c
1 0 +
6 1 -
7 2 -
U b c
1 2 +
6 1 -
7 1 -
Simplifier table…step 3Find value reduct & decision rules:
a0c+
a1c-
a2c-
b2c+
b1c-
U a c
1 0 +
6 1 -
7 2 -
U b c
1 2 +
6 1 -
7 1 -
Example 2 For Case 2 KR-System
U a b c
1 0 2 +
2 0 1 +
3 1 0 +
4 0 1 +
5 1 0 0
6 1 1 0
7 2 1 -
8 0 1 -
9 1 0 -
Attributes of learner’s knowledge : B={a,b}
Attributes of knower’s knowledge : c
Attribute value 0: knower can’t classify this object.
Example 2 For Case 2KR-System
3 concepts of knower’s knowledge:
X0={5,6} ignorance region
X+={1,2,3,4}
X-={7,8,9}
X*=X+ X- competence region5 concepts of learner’s knowledge:
Y0={1}
Y1={2,4,8}
Y2={3,5,9}
YY3={6}={6}
Y4={7}
U a b c
1 0 2 +
2 0 1 +
3 1 0 +
4 0 1 +
5 1 0 0
6 11 11 0
7 2 1 -
8 0 1 -
9 1 0 -
Can the learner be able to discover the ignorance region?
BX+ = Y0= {1}
BX+ = Y0 Y1 Y2 = {1,2,3,4,5,8,9}
X+ is roughly Y-definable, so learner can learn object {1}, not sure about {2,3,4,5,8,9}.
BX- = Y4 = {7}
BX- = Y1 Y2 Y4 = {2,3,4,5,7,8,9}
X- is roughly Y-definable, so learner can learn object {7}, not sure about {2,3,4,5,8,9}.
BX0 = Y3= {6}
BX0 = Y2 Y3 = {3,5,6,9}
X0 is roughly Y-definable, so learner can learn object {6}, not sure about {3,5,9}.
X+={1,2,3,4}
X-={7,8,9}
X*=X+ X- competence region
X0={5,6} ignorance region
Y0={1}
Y1={2,4,8}
Y2={3,5,9}
Y3={6}
Y4={7} Learner can discover the ignorance region X0
The ignorance region influences the learner’s ability to learn?
In this example, the answer is YES, because the ignorance region(X0) is roughly learnable, object 6 is important. Hence, we can prove it:
BX+ = Y0= {1}
BX+ = Y0 Y1 Y2 = {1,2,3,4,5,8,9}
BX- = Y4 = {7}
BX- = Y1 Y2 Y4 = {2,3,4,5,7,8,9}
X*=X+ X- competence region
X0={5,6} ignorance region
X* X* put 6 into X+
X*put 5 into X-
Positive objects
1, 7 1,6,7 1, 7
Border-line objects
2,3,4,5,8,9 2,3,4,5,8,9 2,3,4,5,8,9
Negative objects
6 Ø 6
Object 6 will fall in positive region if knower knows it (move from X0 to X+).
Therefore, X0 affects the learning process.BNB(X*)= BNB(X+) BNB(X-)
Put 6 into X+
X+={1,2,3,4,6}
X-={7,8,9}
X0={5}
BX+ = Y0= {1,6}
BX+ = Y0 Y1 Y2 Y3 = {1,2,3,4,5,2,3,4,5,6,8,98,9}
BX- = Y4 = {7}
BX- = Y1 Y2 Y4 = {2,3,4,52,3,4,5,7,8,98,9}
POSB(X*)= POSB(X+) POSB(X-)
={1,6} {7}
={1,6,7}
X0={5,6} ignorance
region
X+={1,2,3,4}
X-={7,8,9}
X*=X+ X- competence
region
Y0={1}
Y1={2,4,8}
Y2={3,5,9}
Y3={6}
Y4={7}
Calculation Sheet:Positive objects
Simplifier table…step 1Remove inconsistent rows :
U a b c
1 0 2 +
6 1 1 0
7 2 1 -
Object 6 is learnable, therefore learner can discover the ignorance region {4,5} and the ignorance region will affect the learning process.
U a b c
1 0 2 +
2 0 1 +
3 1 0 +
4 0 1 +
5 1 0 0
6 11 11 0
7 2 1 -
8 0 1 -
9 1 0 -
Simplifier table…step 2
{a}
Find attributes reduct :
Delete a:
Delete b:
U a b c
1 0 2 +
6 1 1 0
7 2 1 -U a c
1 0 +
6 1 0
7 2 -
U b c
1 2 +
6 1 0
7 1 -
A is indispensable
Simplifier table…step 3Find value reduct & decision rules:
a0c+
a1c0
a2c-
U a c
1 0 +
6 1 0
7 2 -
Object 6 is learnable, so learner can discover the knower’s ignorance.
Chapter 12 Machine LearningChapter 12 Machine LearningInductive LearningInductive Learning
Presented by: Kai ZhuPresented by: Kai Zhu
Professor: Dr. T.Y. LinProfessor: Dr. T.Y. Lin
Class ID: 117Class ID: 117
In the pervious chapters, we assumed that the set of In the pervious chapters, we assumed that the set of instances U is constant and unchanged during the instances U is constant and unchanged during the
learning process. learning process.
In any real life situations however this is not the In any real life situations however this is not the case and new instances can be added to the set U. case and new instances can be added to the set U.
Example Example
Lets consider the K-R system Lets consider the K-R system given in Example 1 in the given in Example 1 in the
pervious section in chapter 12pervious section in chapter 12
Table 9Table 9
U a b c d e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
4 0 0 1 2 1
5 2 1 0 2 1
6 0 0 1 2 2
7 2 0 0 1 0
8 0 1 2 2 1
9 2 1 0 2 2
10 2 0 0 1 0
Consistent table:Consistent table:
U a b c d e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
7 2 0 0 1 0
8 0 1 2 2 1
10 2 0 0 1 0
Inconsistent table:Inconsistent table:U a b c d e
4 0 0 1 2 1
5 2 1 0 2 1
6 0 0 1 2 2
9 2 1 0 2 2
(e)=card(1,2,3,7,8,10)/card(1,2,3,4,5,6,7,8,9,10)=6/10=0.6
After remove duplicate After remove duplicate and inconsistentand inconsistent
U a b c d e
1 1 2 0 1 1
2 2 0 0 1 0
3 0 1 2 2 1
After computing, get core After computing, get core and reduct values tableand reduct values table
U a b c d e
1 - - - - 1
2 - - - - 0
3 - - - - 1
U a b c d e
1(1) 1 x x x 1
1(2) x 2 x x 1
2(1) 2 x x x 0
2(2) x 0 x x 0
3(1) 0 x x x 1
3(2) x 1 x x 1
3(3) x x 2 x 1
3(4) x x x 2 1
Core tableCore table
Reduct tableReduct table
From reduct table, there are 16 From reduct table, there are 16 Simplified Tables. One of them : Simplified Tables. One of them :
corresponding decision corresponding decision algorithms:algorithms:
U a b c d e
1(1) 1 x x x 1
2(1) 2 x x x 0
3(1) 0 x x x 1
a1a1 e1 e1a2 a2 e0e0a0 a0 e1e1
Table 10Table 10
U a b c d
e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
4 0 0 1 2 1
5 2 1 0 2 1
6 0 0 1 2 2
7 2 0 0 1 0
8 0 1 2 2 1
9 2 1 0 2 2
10 2 0 0 1 0
11 0 1 2 2 1
Consistent table:Consistent table:
Inconsistent table:Inconsistent table:
(e)=card(1,2,3,7,8,10,11)/card(1,2,3,4,5,6,7,8,9,10,11)=7/11=0.636
U a b c d
e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
7 2 0 0 1 0
8 0 1 2 2 1
10 2 0 0 1 0
11 0 1 2 2 1
U a b c d
e
4 0 0 1 2 1
5 2 1 0 2 1
6 0 0 1 2 2
9 2 1 0 2 2
After remove duplicate After remove duplicate and inconsistentand inconsistent
U a b c de
1 1 2 0 1 1
2 2 0 0 1 0
3 0 1 2 2 1
After computing, get core After computing, get core and reduct values tableand reduct values table
U a b c d e
1 - - - - 1
2 - - - - 0
3 - - - - 1
U a b c d e
1(1) 1 x x x 1
1(2) x 2 x x 1
2(1) 2 x x x 0
2(2) x 0 x x 0
3(1) 0 x x x 1
3(2) x 1 x x 1
3(3) x x 2 x 1
3(4) x x x 2 1
Core tableCore table
Reduct tableReduct table
From reduct table, there are 16 From reduct table, there are 16 Simplified Tables. One of them : Simplified Tables. One of them :
corresponding decision corresponding decision algorithms:algorithms:
U a b c d e
1(1) 1 x x x 1
2(1) 2 x x x 0
3(1) 0 x x x 1
a1 a1 e1 e1a2 a2 e0e0a0 a0 e1e1
It is obvious that in table 10 the new instance does It is obvious that in table 10 the new instance does not change the decision algorithm, that means that not change the decision algorithm, that means that
the learned concepts will remain the same.the learned concepts will remain the same.
But the quality of learning But the quality of learning (e) changed. (from 0.6 (e) changed. (from 0.6 to 0.636)to 0.636)
Table 11Table 11
U a b c d
e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
4 0 0 1 2 1
5 2 1 0 2 1
6 0 0 1 2 2
7 2 0 0 1 0
8 0 1 2 2 1
9 2 1 0 2 2
10 2 0 0 1 0
11 1 2 0 1 0
Consistent table:Consistent table:
Inconsistent table:Inconsistent table:
(e)=card(3,7,8,10)/card(1,2,3,4,5,6,7,8,9,10,11)=4/11=0.363
U a b c d
e
3 2 0 0 1 0
7 2 0 0 1 0
8 0 1 2 2 1
10 2 0 0 1 0
U a b c d
e
1 1 2 0 1 1
2 1 2 0 1 1
4 0 0 1 2 1
5 2 1 0 2 1
6 0 0 1 2 2
9 2 1 0 2 2
11 1 2 0 1 0
After remove duplicate After remove duplicate and inconsistentand inconsistent
U a b c de
1 2 0 0 1 0
2 0 1 2 2 1
After computing, get core After computing, get core and reduct values tableand reduct values table
Core tableCore table
Reduct tableReduct table
a b c d
e
- - - - 0
- - - - 1
U a b c d
e
1(1) 2 x x x 0
1(2) x 0 x x 0
1(3) x x 0 x 0
1(4) x x x 1 0
2(1) 0 x x x 1
2(2) x 1 x x 1
2(3) x x 2 x 1
2(4) x x x 2 1
From reduct table, there are 16 From reduct table, there are 16 Simplified Tables. One of them : Simplified Tables. One of them :
corresponding decision corresponding decision algorithms:algorithms:
U a b c d
e
1(1) 2 x x x 0
2(1) 0 x x x 1
a2 a2 e0e0a0 a0 e1e1 (Compare to(Compare to
a1a1 e1 e1a2 a2 e0e0a0 a0 e1)e1)
Table 12Table 12
U a b c d
e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
4 0 0 1 2 1
5 2 1 0 2 1
6 0 0 1 2 2
7 2 0 0 1 0
8 0 1 2 2 1
9 2 1 0 2 2
10 2 0 0 1 0
11 1 0 0 1 3
Consistent table:Consistent table:
Inconsistent table:Inconsistent table:
(e)=card(1,2,3,7,8,10,11)/card(1,2,3,4,5,6,7,8,9,10)=6/10=0.636
U a b c d
e
1 1 2 0 1 1
2 1 2 0 1 1
3 2 0 0 1 0
7 2 0 0 1 0
8 0 1 2 2 1
10 2 0 0 1 0
11 1 0 0 1 3
U a b c d
e
4 0 0 1 2 1
5 2 1 0 2 1
6 0 0 1 2 2
9 2 1 0 2 2
After remove duplicate After remove duplicate and inconsistentand inconsistent
U a b c de
1 1 2 0 1 1
2 2 0 0 1 0
3 0 1 2 2 1
4 1 0 0 1 3
After computing, get core After computing, get core and reduct values tableand reduct values table
Core tableCore table
Reduct tableReduct table
U a b c d
e
1 - 2 - - 1
2 2 - - - 0
3 - - - - 1
4 1 0 - - 3
U a b c d
e
1(1) x 2 x x 1
2(1) 2 x x x 0
3(1) 0 x x x 1
3(2) x 1 x x 1
3(3) x x 2 x 1
3(4) x x x 2 1
4(1) 1 0 x x 3
From reduct table, there are 4 From reduct table, there are 4 Simplified Tables. One of them : Simplified Tables. One of them :
corresponding decision corresponding decision algorithms:algorithms:
U a b c d
e
1(1) x 2 x x 1
2(1) 2 x x x 0
3(1) 0 x x x 1
4(1) 1 0 x x 3b2 b2 e1e1a2 a2 e0e0a0a0e1e1
a1b0a1b0e3e3
(Compare to(Compare toa1a1 e1 e1a2 a2 e0e0a0 a0 e1)e1)
To sum up, as we have seen from the above To sum up, as we have seen from the above examples, adding a new instance to the universe examples, adding a new instance to the universe
we could face three possibilities:we could face three possibilities:
1. The new instance confirms actual knowledge. 1. The new instance confirms actual knowledge. (table10)(table10)
2. The new instance contradicts the actual 2. The new instance contradicts the actual knowledge. (table11)knowledge. (table11)
3. The new instance is a completely new case.3. The new instance is a completely new case.
(table12)(table12)
Thank youThank you